Bertrand's Box Paradox

There are three boxes: one with two gold coins, one with two silver coins, one with one of each. You pick a box at random and draw one coin, which turns out to be gold. What's the probability the other coin in that box is also gold? The tempting answer is 1/2 — it must be either the gold-gold box or the mixed box, and those seem equally likely. That answer is wrong, and the reason why is the whole point of the paradox.

The mistake is treating "two remaining possibilities" as automatically equally probable once you've conditioned on drawing gold. Before drawing, each box has equal 1/3 odds. But a gold draw is twice as likely to have come from the gold-gold box (which offers two chances to draw gold) as from the mixed box (which offers only one). Counting individual coins rather than boxes: there are three gold coins total, two of them sitting in the gold-gold box and one in the mixed box. Given that you drew a gold coin, the probability it came from the gold-gold box is $2/3$, not $1/2$ — so the probability the other coin is also gold is $2/3$.